Aluminium and Silver : PS Archive
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# Aluminium and Silver

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VP
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09 May 2009, 10:23
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A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is 2 mm thick and its diameter is 15 mm). The weight of the coin is 30 grams and the volume of aluminum in the alloy equals that of silver. What will be the weight of the coin measuring 1 x 30 mm made of pure aluminum if silver is twice heavier than aluminium?

* 36 grams
* 40 grams
* 42 grams
* 48 grams
* 50 grams
CEO
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09 May 2009, 13:29
Let's move from one coin to another:

1) m1 = 30 g.
2) decrease in thick from 2mm to 1mm: m = 30 * 1/2
3) increase in diameter from 15mm to 30mm: m= 30 * 1/2 * 2^2
4) substitute 1/2 of silver to 1/2 of aluminum: it reduces weight by 3/2 times: m = 30 * 1/2 * 4 * 2/3 = 40 g.
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VP
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09 May 2009, 15:12
walker wrote:
Let's move from one coin to another:

1) m1 = 30 g.
2) decrease in thick from 2mm to 1mm: m = 30 * 1/2
3) increase in diameter from 15mm to 30mm: m= 30 * 1/2 * 2^2
4) substitute 1/2 of silver to 1/2 of aluminum: it reduces weight by 3/2 times: m = 30 * 1/2 * 4 * 2/3 = 40 g.

You make it look so simple but I am afraid I did not get the tail or head of what you tried to explain. How are you relating weight thickness and diameter?

In 2 Are you saying that as thickness decreases weight also decreases and hence divide by 2??

Diameter increased by 2 but why are you multiplying by 2^2 rather than 2? Is it because to account for 2 sides of coin?

What really perplexed me is this Sentence

The weight of the coin is 30 grams and the volume of aluminum in the alloy equals that of silver.

Can you explain why the weight reduces by 3/2 times?

It is talking about weight of the coin and then says volume of AL and Si are same. Its not like they are 15 gms each.

Here is the OE and OA

Denote S the weight of silver and A the weight of aluminum in the first coin. We can compose an equation: S + A = 30 or 2A + A = 30. Thus, the aluminum half of the coin weighs 10 grams. If the coin were made of pure aluminum, the second half of the coin would also weigh 10 grams and the whole coin would weigh 20 grams.

If we look at the proportions of the second coin, we will see that it is twice as large as the first coin (volume2 = $$1(\frac{30}{2})^2 \pi$$ ; volume1 = $$2(\frac{15}{2})^2 \pi)$$ . The second coin were made of pure aluminum would be twice as heavy as the first coin made of pure aluminum. The answer to the question is therefore 20*2 = 40 grams.

While the OE makes a lot of sense in the first part, I dont understand what kind of an object are they considering the coin to be? Sphere?
CEO
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10 May 2009, 06:00
There is a direct way to solve it, as you've posted here: to write a formula of weight for first coin and second coin. After that, express weight of first coin through weight of second coin.

I used indirect way. I actually transform our first coin to second coin by means of a few steps as a blacksmith. It seems obvious for me that if we change the thickness of a coin from 2mm to 1mm we have 1/2 m (or 2 coins instead). The same with radius, changing the radius of a coin from 15mm to 30mm we increase area of the coin by 4 and volume (weight) by the same factor. In the case of [Al][Ag] coin, it has weight that equals to [Al][Al][Al] coin, or coin with volume that 3/2 times more. So, we need use 2/3. Anyway, it was the fastest way for me, but if it is not so obvious or fast for you, it would be better to use the direct way.
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Manager
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11 May 2009, 05:22
Volume of the alloy= Pi * r^2 *h
= 225pi/2
1. Since volume of silver= volume of alluminium= 225pi/4

2. Also, Silver is twice heavier than aluminium
=> Aluminium weight=10gm and Silver wt=20 gm

So when we make an aluminium coin measuring 1 x 30 mm (=> volume= 225pi), the weight will be four times the volume of the original alloy i.e 40gm (from 1).
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17 Oct 2009, 07:07
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The V of the second 1x30mm coin 15^2pi*30/(7.5^2*pi*2)=60
S=2A and the V of S = the V of A so the weight of the second coin made of pure A is 60/3*2=40
Re: Aluminium and Silver   [#permalink] 17 Oct 2009, 07:07
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